1 / 13

The Mole and Molar Mass

The Mole and Molar Mass. Moles are the currency of chemistry 1 mole = 6.02 x 10 23 particles = Avogadro’s number Molar mass of an element = mass of one mole of that element Molar mass = atomic mass of element in grams - Average mass of one atom of C = 12.01 amu

lars-leonard
Download Presentation

The Mole and Molar Mass

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Mole and Molar Mass • Moles are the currency of chemistry • 1 mole = 6.02 x 1023 particles = Avogadro’s number • Molar mass of an element = mass of one mole of that element • Molar mass = atomic mass of element in grams - Average mass of one atom of C = 12.01 amu - Average mass of one mole of C = 12.01 g • Molar mass of a compound = sum of molar masses of each atom in the compound • Example: What is the molar mass of MgCl2? Molar mass of Mg = 24.31 g, Molar mass of Cl = 35.45 g Molar mass of MgCl2 = 24.31 g + (2 x 35.45 g) = 95.21 g • *Note: use as many sig. figs. as given in the atomic masses (don’t round molar masses to 1/10 of a gram)

  2. Calculations using Molar Mass • Molar mass is used to convert between moles and grams of a substance • Chemical equations are balanced with atoms, and moles are an extension of that (they also balance) • Quantities of reactants and products are measured by mass (grams) in the lab, but we use moles to balance equations • Example 1: How many grams of H2O are in 0.100 mol H2O? Molar mass of H2O = (2 x 1.008 g) + 16.00 g = 18.02 g Grams H2O = 0.100 mol x (18.02 g/1 mol) = 1.80 g H2O • Example 2: How many moles are in 0.250 g of HCl? Molar mass of HCl = 1.008 g + 35.45 g = 36.46 g Mol HCl = 0.250 g x (1 mol/36.46 g) = 6.86 x 10-3 mol HCl

  3. Percent Composition • Molecular formula tells you how many moles of each type of atom are in one mole of a compound • Using molar mass, you can calculate the % by mass of each element in a compound • Example: Calculate the % composition by mass of CO2 Molar mass of CO2 = 12.01 g + (2 x 16.00 g) = 44.01 g %C = (12.01 g/44.01 g) x 100% = 27.29% %O = (2 x 16.00 g/44.01 g) x 100% = 72.71%

  4. Empirical Formulas • Empirical formulas give the lowest whole-number ratio of atoms in a compound • Molecular formula = multiple of empirical formula • Example: empirical formula for glucose = CH2O molecular formula = C6H12O6 • Can calculate empirical formula from masses or from % composition by mass • Example: Calculate empirical formula for compound with 8.0 g C and 2.0 g H 8.0 g C x (1 mol/12.01 g) = 0.6661 mol C 2.0 g H x (1 mol/1.008 g) = 1.984 mol H 0.6661 mol C/0.6661 mol = 1.000 C = 1 C 1.984 mol H/0.6661 mol = 2.979 H = 3 H’s Empirical formula = CH3

  5. Mole Relationships and Mole Calculations • Atoms, moles and mass are conserved in a chemical reaction (shown in balanced equation) • Using chemical formulas, molar masses and mole ratios (from balanced chemical equation), amounts of products and reactants can be calculated • Example: For C3H8 + 5O2 3CO2 + 4H2O How many moles of O2 are needed to produce 2.0 moles of H2O? • Have moles H2O, want moles O2 • 5 mol O2 = 4 mol H2O • 2.0 mol H2O x (5 mol O2 /4 mol H2O) = 2.5 mol O2

  6. Mole Relationships and Mass Calculations Example: C3H8 + 5O2 3CO2 + 4H2O How many grams of CO2 are produced when 10.0 g of C3H8 are consumed? 1. Have grams C3H8, want grams CO2 (g C3H8  mol C3H8  mol CO2  g CO2) 2. Molar mass C3H8 = (3 x 12.01 g) + (8 x 1.008 g) = 44.094 g Mol C3H8 = 10.0g x (1 mol/44.094 g) = 0.2268 mol 3. Mol CO2 = 0.2268 mol C3H8 x (3 mol CO2/1 mol C3H8) = 0.6804 mol 4. Molar mass CO2 = 12.01 g + (2 x 16.00 g) = 44.01 g Grams CO2 = 0.6804 mol CO2 x (44.01 g/1 mol) = 29.9 g

  7. Percent Yield • Theoretical (or ideal) yield is calculated from balanced chemical equation and molar masses • Actual yield = what you got (usually < ideal) • % yield = (actual yield/theoretical yield) x 100% Example: 2Fe + 3S  Fe2S3 What is the % yield if 1.00 g Fe makes 1.50 g Fe2S3? 1. Calculate theoretical yield: (g Fe  mol Fe  mol Fe2S3  g Fe2S3) 1.00 g Fe x (1 mol/55.85 g) = 0.0179 mol Fe 0.0179 mol Fe x (1 mol Fe2S3/2 mol Fe) = 0.00900 mol Fe2S3 Molar mass Fe2S3 = (2 x 55.85 g) + (3 x 32.06 g) = 207.88 g 0.00900 mol Fe2S3 x (207.88 g/1 mol) = 1.87 g Fe2S3 2. % yield = (1.50 g Fe2S3 /1.871 g Fe2S3) x 100% = 80.2%

  8. Limiting Reactant • Limiting Reactant = reactant that is completely used up • Example: If 1.0 mol of Fe are reacted with 3.0 mol of S, which is the limiting reactant? - Since 2.0 mol Fe are required to react with 3.0 mol of S, then 1.0 mol of Fe can only react with 1.5 mol of S - So, all the Fe is use up and 1.5 mol of S is leftover - Fe is the limiting reactant

  9. Equilibrium Constants • For reversible reactions at equilibrium: - Rate of forward reaction = rate of reverse reaction - Amounts of reactants and products are constant (but not necessarily equal) • Ratio of products to reactants = equilibrium constant (Keq) • For a reaction aA + bB  cC + dD Keq = [C]c [D]d/[A]a [B]b, where [A] = concentration of A • Example: for 2H2 + O2 2H2O Keq = [H2O]2/[H2]2 [O2] • If Keq > 1, equilibrium favors products If Keq < 1, equilibrium favors reactants

More Related